Question

If we were to have a box of marbles containing 2 red, 2 black, 2 blue, and 2 yellow, and we drew out a marble, we would then have the choice of keeping the marble out or putting it back. Each choice affects the odds (probability) of subsequent draws differently. If we draw out a red marble and put it back, this is known as with replacement, and the odds of subsequent draws are the same is for the first draw. If we draw a red marble and keep it out, known as drawing without replacement, the total number of marbles decreases as does the specific color of marble chosen. For example, if we draw two marbles from our box of marbles, what are the chances the first draw will be red?

            p(red on 1^st draw) =

If the first draw was a red marble, what are the odds of the second draw being a red marble if we didn’t replace the 1^st red marble we drew out?

            p(red on 2^nd draw) =

Combine these draws; what is the probability that the first and second draws (without replacement) will be red?

Answer 1

p(red on 1^st draw) = 2/8=1/4 (As there are 2 red marbles out of 8)

P( red on 2^nd draw) = 1/7 (as there remains 1 red out of 7 remaining marbles)

probability that the first and second draws (without replacement) will be red

=(1/4)*(1/7)=1/28